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A particle moves along a straight line according to the law $s=\frac{1}{3} t^3-3 t^2+9 t+17$, where $s$ is in metre and $t$ is in second. Its velocity decreases in
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Verified Answer
The correct answer is:
$0 < t < 3$
We have,
$$
\begin{aligned}
& S=\frac{1}{3} t^3-3 t^2+9 t+17 \\
& V=\frac{d s}{d t}=t^2-6 t+9 \\
& V=t^2-6 t+9 \\
& \frac{d V}{d t}=2 t-6
\end{aligned}
$$
Velocity is decreasing, if $\frac{d V}{d t} < 0$
$$
\begin{array}{ll}
\therefore \quad & 2 t-6 < 0 \\
& 0 < t < 3$ [\because t>0]
\end{array}
$$
$$
\begin{aligned}
& S=\frac{1}{3} t^3-3 t^2+9 t+17 \\
& V=\frac{d s}{d t}=t^2-6 t+9 \\
& V=t^2-6 t+9 \\
& \frac{d V}{d t}=2 t-6
\end{aligned}
$$
Velocity is decreasing, if $\frac{d V}{d t} < 0$
$$
\begin{array}{ll}
\therefore \quad & 2 t-6 < 0 \\
& 0 < t < 3$ [\because t>0]
\end{array}
$$
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